Determination of abstract groups of small order

Determination of groups of order 8

Let $G$ be a group with $\#G=8$ . If $G$ is Abelian then, by the classification of finite Abelian groups, $G$ is one of:
$⟨a:a^8=1⟩$
$⟨a,b:a^4=b^2=1,ab=ba⟩$
$⟨a,b,c:a^2=b^2=c^2=1,ab=ba,ac=ca,bc=cb⟩$

Assume $G$ is not Abelian. Let $Z$ be the center of $G$. $G$ is a 2-group so $Z\ne 1$ and, by assumption, $Z\ne \mathrm{G}$ , so $G/Z$ has order 2 or 4. $G/Z$ cannot by cyclic and the only group of order 2 or 4 which is not cyclic is $⟨b,c:b^2=c^2=1,bc=cb⟩$ . This has order 4 so $Z$ has order 2 and must be $⟨a:a^2=1⟩$ . We apply central extension theory. Let $G/Z=⟨Zb,Zc⟩$ with $b^2,c^2,[b,c]\in Z$ . Computing the homology group of $G/Z$ , we find that the only relation on $b^2,c^2,[b,c]$ is ${[b,c]}^2=1$ which is vacuous in this case. G is not Abelian so $[b,c]\ne 1$ . We get four possibilities:

$\begin{array}{}{b}^{2}11aa\\ c^{2}1a1a\\ [b,c]aaaa\end{array}$

We now determine which of these columns will produce isomorphic groups. First, as is easy to compute, multiplying $b$ or $c$ by $a$ has no effect on the values of $b^2,c^2,[b,c]$ . So there are no equivalent columns due to the choice of coset representatives. Second, the automorpism group of $Z$ is trivial so there are no equivalent colums due to the choice of generator for $Z$ . Third, the automorphism group of $⟨b,c:b^2=c^2=1,bc=cb⟩$ is generated by $(b,c)\to (bc,c)$ and $(b,c)\to (c,bc)$ . The effect of the first of these on the values of $b^2,c^2,[b,c]$ is to interchange the first and third columns of the table. The effect of the second is to interchange the first and second columns. Therefore the first three columns yield isomorphic groups and the fourth column yields a distict group. We get two sets of generators and relations based on the second and fourth columns:
$⟨a,b,c:a^2=1,b^2=a,c^2=1,ab=ba,ac=ca,bc=cba⟩$
$⟨a,b,c:a^2=1,b^2=a,c^2=a,ab=ba,ac=ca,bc=cba⟩$
These simplify to:
$⟨b,c:b^4=c^2=1,bc=c{b}^{-1}⟩$
$⟨b,c:b^4=1,c^2=b^2,bc=c{b}^{-1}⟩$
Or:
$⟨a,b:a^4=b^2=1,ab=b{a}^{-1}⟩$
$⟨a,b:a^4=1,b^2=a^2,ab=b{a}^{-1}⟩$

The total number of nonisomorphic groups is 5.